# How many circles fit in the same two dimensional orbit?

I'm trying to figure out how many circles/planets can fit in the same two dimensional orbit. All of the planets and orbits are perfect circles but the planets size and amount can vary.

This is for some software I'm writing, here are likely problems I need to solve for:

orbit_circumference = 500


How many fit?

orbit_circumference = 500
total_planets = 16


If we know 4 of the planets have the radius of 10 what is the max radius for the other 12?

Just hoping someone can get me pointed in the right direction!

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If you take $n$ equally-spaced points on a circle of radius $R$, the distance between neighbouring points is $2 R \sin(\pi/n)$. So if you want $n$ circular "planets" with centres on the circle of radius $R$, in order not to overlap the planets must have radius at most $R \sin(\pi/n)$. Conversely, if the planets have radius $r$, the number of planets must be at most $\dfrac{\pi}{\arcsin(r/R)}$.

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If all the $n$ planets have the same radius, they each occupy $\frac {2\pi}n$ radians as seen from the center. To a first approximation, valid if there are lots of planets, the planet diameter is the circumference of the orbit divided by $n$. The approximation comes because this takes the points of tangency of the planets to be at opposite ends of a diameter, which is not quite true.

Again if the planets are the same diameter, draw the two tangents from the center of the orbit to one planet. Also draw the line from the center of the orbit to the center of the planet and the two radii from the center of the planet to the points of tangency. You have two right triangles. The hypotenuse is the line to the center of the planet, of length $R$, the radius of the orbit. The radius of the planet is $r$. The length of the tangent is $\sqrt {R^2-r^2}$ and the angle at the center of the orbit is half what we had above, so we get $\sin \frac \pi n=\frac rR$ or $r=R \sin \frac \pi n$. The approximation above is the small angle $\sin \frac \pi n \approx \frac \pi n$

For your last question, if you are given one population of planets of one size and the orbit radius, you can calculate how much of the angle they take up and therefore what is left for the others. There are small corrections because the tangency points move around, but they should not be important.

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Let a small circle of radius $r$ have its center on the circumference of a large circle of radius $R$. Then seen from the center of the large circle, the small circle subtends an angle $2\phi$ where $\sin\phi={r\over R}$.

Therefore, if a collection of radii $r_i>0$ $\ (1\leq i\leq n)$ is given, $n$ circles of these radii can be placed on the circumference of a large circle of radius $R$ without overlap iff $$\sum_{i=1}^n \arcsin{r_i\over R}\leq\pi\ .$$ To solve your last problem, put $R:={500\over2\pi}$ and solve the equation $$12\arcsin{r\over R}=\pi-4\arcsin{10\over R}$$ for the unknown $r$. You only need a "technical" pocket calculator for this.

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